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Mathematics
OpenStudy (anonymous):

is f'(x) of f(x)=3x^2 e^-2 x^2 -2?

OpenStudy (anonymous):

my bad, it is e^-x

OpenStudy (anonymous):

too bad you cant edit

OpenStudy (anonymous):

s0 (3x^2)(e^-x)(x^2) -2 ??

OpenStudy (anonymous):

(6xe^-x)-3x^2 e^-x use the product rule, u will get the answer

OpenStudy (anonymous):

F(x) = (3x^2)(e^-x) What is f'(x)?

OpenStudy (anonymous):

I got x^2 - 2

OpenStudy (anonymous):

d{(3x^2)(e^-x)}/dx= 3x^2 d(e^-x)/dx + (e^-x) d(3x^2)/dx = (6xe^-x)-3x^2 e^-x

OpenStudy (anonymous):

now factor out -3x

OpenStudy (anonymous):

and e^-x

OpenStudy (anonymous):

yep, it will be: (3xe^-x)(2-x)

OpenStudy (anonymous):

So where am i goofing up? (3x^2)(-1e^-x) + (e^-x)(6x)

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

so -3x^2 + 6x after e^-x is factored out

OpenStudy (anonymous):

also take common 3x

OpenStudy (anonymous):

so we are left with -x+2?

OpenStudy (anonymous):

yep, thats what i wrote before ^^

OpenStudy (anonymous):

ok, I was only a little backwards. thanks

OpenStudy (anonymous):

welcome..

OpenStudy (anonymous):

and f''(x) is 1...

OpenStudy (anonymous):

i have to check, but i dont think so.. again u have to apply product rule in f'(x) to find out f"(x)

OpenStudy (anonymous):

I thought second derivitive was the derivitive of the first?

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